Question: What is the area of the portion of the circle defined by $x^2-12x+y^2=28$ that lies above the $x$-axis and to the right of the line $y=6-x$?
Answer: Completing the square, the equation of the circle can be rewritten in the form \[
(x^2-12x +36) +y^2=64,
\]or $(x-6)^2 +y^2 =8^2.$ The center of this circle is $(6,0)$, so both the $x$-axis and the line $y=6-x$ pass through the center of the circle: [asy]
size(8cm);
void axes(real x0, real x1, real y0, real y1)
{
    draw((x0,0)--(x1,0),EndArrow);
    draw((0,y0)--(0,y1),EndArrow);
    label("$x$",(x1,0),E);
    label("$y$",(0,y1),N);
    for (int i=floor(x0)+1; i<x1; ++i)
        draw((i,.1)--(i,-.1));
    for (int i=floor(y0)+1; i<y1; ++i)
        draw((.1,i)--(-.1,i));
}
void e(real a, real b, real h, real k)
{
    draw(shift((h,k))*scale(a,b)*unitcircle);
}
filldraw(arc((6,0),8,0,135)--(6-4*sqrt(2),4*sqrt(2))--(6,0)--cycle,rgb(0.8,0.8,0.8));
axes(-5,18,-10,10);
e(8, 8, 6, 0);
real f(real x) { return 6-x; }
draw(graph(f,-3,14),Arrows);
dot((6,0));
label("$y=6-x$",(14,-8),E);
label("$(6,0)$",(6,0),dir(235)*1.3);
[/asy] Since the line $y=6-x$ has slope $-1,$ it makes an angle of $135^\circ$ with the positive $x-$axis, so the desired region makes up $\frac{135^\circ}{360^\circ} = \frac{3}{8}$ of the circle.  The radius of the circle is $\sqrt{64} = 8$, so the circle has area $8^2\pi = 64\pi$. Therefore, the desired area is $64\pi\cdot\frac{3}{8}=\boxed{24 \pi}$.